Divisibility of $a_{24}$ by 7. ($a_n=\underbrace{999\cdots9 }_{n \text{ times}})$

Question

Question: By which number is $a_{24}$ divisible by?
Where $a_n=\underbrace{999\cdots9 }_{n \text{ times}}$

The solution says the answer is $7$. Here's what is given:

$$a_{24}=\underbrace{999\cdots9 }_{24 \text{ times}}$$ $$=9(\underbrace{\underline{111} \ \ \underline{111}\ \ \underline{111} \ \cdots \ \ \underline{111})}_{8 \text{ similar sets}}$$ Now differences of each set is $0$. Hence $a_{24}$ is divisible by $7$.

Now what I don't understand is what are they implying when they say "difference of each set is $0$" . Also , why does this imply that the number is entirely divisible by $7$?

Also I know the divisibility rule of $7$ to be: Double the last digit, subtract the obtained number from whatever remains after removing the last digit and then check if the final number obtained is divisible by 7.

This process can go lengthy for this question here. Is there any way to solve it quicker?

Use this duplicate for an easy proof. Note that $24=4\cdot 6$. — Dietrich Burde, May 25 '20 at 19:16
@DietrichBurde , thanks for the link but I am totally unfamiliar with the stuff used to answer the question. — sai-kartik, May 25 '20 at 19:19
Since $7$ divides $1001$, another rule for divisibility by $7$ Involves subtract alternating $3$ digits; for example, $15001$ is divisible by $7$ because $15-1 $ is — J. W. Tanner, May 25 '20 at 19:21
"but I am totally unfamiliar with the stuff used" That seems like a good place to start then. Prove or learn the proof that $x^{p-1}\equiv 1\pmod{p}$ for every prime $p$ and $x$ coprime to $p$. Then, use the result. — JMoravitz, May 25 '20 at 19:21
"By which number"? Yes, it's divisible by $7$, but also by $3$, $9$, $11$, $13$, and many others. — Robert Israel, May 25 '20 at 19:25
@RobertIsrael but the options were stated: 24, 12 , 7 and 31 so I bothered to ask only for 7 — sai-kartik, May 25 '20 at 19:27
Also I appreciate everyone helping me here and sorry for not mentioning this sooner, but I'm still a student in 12th grade and I haven't learnt any of this stuff. However, I now know where to begin — sai-kartik, May 25 '20 at 19:28

score 2 · Accepted Answer · answered May 25 '20 at 19:34

To address what you don’t understand about the solution given, you should become aware that there is another rule for divisibility by $7$ besides the one you mentioned. This rule is to alternately add and subtract $3$-digit chunks of the number starting with the last $3$ digits and testing whether the result is divisible by $7$. For example, $7003010$ is divisible by $7$ because $10-3+7$ is. This rule works because $7$ divides $1001.$ (By the way, it works for $11$ and $13$ too.) Using this rule, it becomes obvious that any number written as a string of $n$ $1$s, where $n$ is a multiple of $6$, is divisible by $7$.

score 1 · Answer 2 · answered May 25 '20 at 19:23

1

Observe that $$a_n=\dfrac{10^n-1}{10-1},n\ge1$$

Now as $(10,7)=1$ and $\phi(7)=6$

$7$ will definitely divide $10^n-1$ if $6$ divides $n$

Again as $(10-1,7)=1,7$ will divide $a_n$ if $6$ divides $n$

answered May 25 '20 at 19:23

lab bhattacharjee

274,582

Could you please define $(10,7)$ and $\phi(7)$ – sai-kartik May 25 '20 at 19:24
1

@sai-kartik $(10,7)$ is alternate notation for $\gcd(10,7)$, the greatest common divisor. Then, $\phi(7)$ is the number of natural numbers less than $7$ coprime to $7$ and is called Euler's Totient Function. – JMoravitz May 25 '20 at 19:25
1

The answer brings these up since there is a generalization of Fermat's Little Theorem which says that $a^{p-1}\equiv 1\pmod{p}$ for any prime $p$ and $a$ coprime to $p$ which is called Euler's Theorem which says more generally that if $\gcd(a,n)=1$ then $a^{\phi(n)}\equiv 1\pmod{n}$. Of course, the usual Fermat's Little Theorem is sufficient for this problem since $7$ is prime. – JMoravitz May 25 '20 at 19:29
@JMoravitz Thank you for taking time to answer this question. (See last comment on the question) – sai-kartik May 25 '20 at 19:31

fleablood · Answer 3 · 2020-05-25T20:53:54.867

I think the text is a assuming a different divisible by $7$ rule. (The rule that J.W. Tanner pointed out in the comments.)

If you have a number that is $3k$ digits long $N= a_kb_kc_ka_{k-1}b_{k-1}c_{k_1}.....a_2b_2c_2a_1b_1c_1$ (you can add zeros to make it $3k$ digits long) you do

$M = a_kb_kc_k - a_{k-1}b_{k_1}c_{k_1} + ...... $ and seeing if $7$ divides $M$.

For example:

To find out if $42405231722$ is divisible by $7$ we take $-42+405 - 231+722 = 363 + 491=854$ and $854$ divided by $7$ is $122$ with no remainder so it is divisible by $7$.

So for $N= 111,111,111,111,111,111,111,111$ we take $111-111 + 111-111+111-111 + 111 -111 = 0$ so $N$ is divisible by $7$.

Why does the rule work?

Well $1001 = 7*143$ so $abcdef = abc*1000 + def = abc*1001 + def-abc = 7(abc*143) + (def-abc)$ so $abcdef$ will have the same remainder when divided by $7$ and $def-abc$.

So $-42+405 - 231+722$ has the same remainder as $42405 + 231722$ what has the same remainder as $-42405*1000 + 231722$ which has the same remainder as $42405*10^6 + 231722=42405231722$.

.....

All said and done this is not a good way to do it.

Better to note that $10^{ab} - 1 = (10^a-1)(10^{ab-b} + 10^{ab-2b} + .... + 10^b + 1)$ so $10^a -1|10^{ab}-1$ and then if $a$ is a prime other than $2$ and $5$ then $a|10^a-1$ and $a|10^{ab}-1$.

So $7|10^6-1 = 999,999$ and so $7|10^{24}-1 = (10^6-)(10^{18} + 10^{12} + 10^6 + 1)$.

This rule has been pointed to by @J.W. Tanner. However, I really appreciate the explanation you have given. — sai-kartik, May 25 '20 at 20:36
Hmm.... I meant $1001 = 7*143$ but don't know how I made such a typo. — fleablood, May 25 '20 at 20:53
Yes, I was explaining why the rule worked, but you are right, I should give J.W. Tanner the credit he/she deserves. — fleablood, May 25 '20 at 20:54

Divisibility of $a_{24}$ by 7. ($a_n=\underbrace{999\cdots9 }_{n \text{ times}})$

3 Answers3